Question Number 222684 by alvan545 last updated on 05/Jul/25
Commented by alvan545 last updated on 05/Jul/25
Answered by mr W last updated on 05/Jul/25
Commented by mr W last updated on 05/Jul/25
$$\Delta{ABC}\:\sim\:\Delta{EBA} \\ $$$$\frac{{c}}{{b}}=\frac{{b}+{a}+{d}}{{c}} \\ $$$$\Rightarrow{bd}={c}^{\mathrm{2}} −{b}\left({a}+{b}\right) \\ $$$${ac}^{\mathrm{2}} +{bd}^{\mathrm{2}} =\left({a}+{b}\right)\left({x}^{\mathrm{2}} +{ab}\right)\:\:\:\:\:\left({Stewart}'{s}\:{theorem}\right) \\ $$$${x}^{\mathrm{2}} =\frac{{ac}^{\mathrm{2}} +{bd}^{\mathrm{2}} }{{a}+{b}}−{ab} \\ $$$${x}^{\mathrm{2}} =\frac{{abc}^{\mathrm{2}} +{c}^{\mathrm{4}} +{b}^{\mathrm{2}} \left({a}+{b}\right)^{\mathrm{2}} −\mathrm{2}{b}\left({a}+{b}\right){c}^{\mathrm{2}} −{a}\left({a}+{b}\right){b}^{\mathrm{2}} }{{b}\left({a}+{b}\right)} \\ $$$${x}^{\mathrm{2}} =\frac{\left({c}^{\mathrm{2}} −{b}^{\mathrm{2}} \right){c}^{\mathrm{2}} −{b}\left({a}+{b}\right)\left({c}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)}{{b}\left({a}+{b}\right)} \\ $$$${x}^{\mathrm{2}} =\frac{\left({c}^{\mathrm{2}} −{b}\left({a}+{b}\right)\right)\left({c}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)}{{b}\left({a}+{b}\right)} \\ $$$$\Rightarrow{x}=\sqrt{\frac{\left({c}^{\mathrm{2}} −{b}\left({a}+{b}\right)\right)\left({c}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)}{{b}\left({a}+{b}\right)}}\:\:\:\checkmark \\ $$
Commented by alvan545 last updated on 05/Jul/25
$$\mathrm{how}\:\mathrm{to}\:\mathrm{insert}\:\mathrm{image}\:\mathrm{in}\:\mathrm{comment}\:\mathrm{section}? \\ $$
Commented by alvan545 last updated on 05/Jul/25
$$\mathrm{100}\checkmark,\:\mathrm{thankyou}\:\mathrm{Sir} \\ $$
Commented by mr W last updated on 05/Jul/25
$${you}\:{can}\:{only}\:{attach}\:{an}\:{image}\:{as}\:{new} \\ $$$${comment}.\:{but}\:{you}\:{can}\:{not}\:{insert}\:{an} \\ $$$${image}\:{into}\:{an}\:{other}\:{comment}. \\ $$
Commented by alvan545 last updated on 05/Jul/25
$$\mathrm{thankyou},\mathrm{Sir}\: \\ $$